计数基本原理的重要性是什么？

概率定义了该事件的所有可能结果/结果中该事件发生的度量。事件的概率始终介于 0 到 1 之间。当概率为零时，表示没有机会发生有利/可喜的结果。而另一方面，当概率为 1 时，它表示肯定每次可爱的结果都会作为事件的结果出现。因此，在考虑概率场景时，数字知识变得非常重要。

计数的基本原理

计数的基本概率表明，如果存在一个概率场景，其中有 x ₁ 、 x ₂ 、 x ₃ ... x _n实体对象，每个实体对象都有 y ₁ 、 y ₂ 、 y ₃ ... y _n可供每个实体选择，那么方式的数量，

方式 = y ₁ × y ₂ × y ₃ × … × y _n

在这里，每个可用实体的选择相乘以获得选择的总方式。

计数基本原理的重要性是什么？

回答：

The Fundamental Principle of Counting is essential as it has the below characteristics:

• Fundamental Principle of Counting  helps to determine how selection is done on the basis of available choices.
• Fundamental Principle of Counting simplifies the approach of selection considering all the possible choices and their combinations for calculating the Probability.
• The Fundamental Principle of Counting is one such vital part of Probability which deals with the knowledge of numbers and there much-needed use when considered from the knowledge of Mathematics.

Example of Fundamental Counting Principle Problem, Consider Seema has 2 blue pens, 2 black and 2 red pens. In how many ways can she select one pen of each kind,

Then pairing can take place as follows:

(B₁ b₁ r₁), (B₁ b₁ r₂), (B₁ b₂ r₁), (B₁ b₂ r₂), (B₂ b₁ r₁), (B₂ b₁ r₂), (B₂ b₂ r₁), (B₂ b₂ r₂)

The total number of ways of choosing this pairing using Counting Principle Problems,

• Choices available for  blue pens = 2
• Choices available for black pens = 2
• Choices available for red pens = 2

Total number of ways: 2 × 2 × 2 = 8

类似问题

问题 1：基本计数原理是否始终适用于计数问题。

回答：

Yes, Fundamental Counting Principle always holds for Counting Problems.

问题 2：考虑一位老师，他有 1 支黑笔和 2 支红笔。她有多少种方法可以选择每种钢笔。

解决方案：

Then pairing can take place as follows,

(B₁ R₁), (B₁ R₂)

• Choices available for black pens = 1
• Choices available for red pens = 2

Total number of ways: 1 × 2 = 2

问题 3：假设一个男孩可以选择 2 杯茶和 2 杯咖啡。他有多少种方法可以选择每种杯子。

解决方案：

Then pairing can take place as follows,

(T₁ C₁), (T₁ C₂), (T₂ C₁), (T₂ C₂) 

• Choices available for tea = 2
• Choices available for coffee = 2

Total number of ways: 2 × 2  = 4

问题 4：假设一个孩子有 3 个棒棒糖和 2 个太妃糖。她有多少种方法可以选择每种糖果。

回答：

Then pairing can take place as follows,

(L₁ T₁), (L₁ T₂), (L₂ T₁), (L₂ T₂), (L₃ T₁), (L₃ T₂)

• Choices available for black pens = 3
• Choices available for red pens = 2

Total number of ways: 3 × 2 = 6

问题 5：假设一个女孩有 3 个黑色上衣和 2 个蓝色下衣。她有多少种方法可以选择上装和下装的连衣裙组合。

解决方案：

Then pairing can take place as follows:

(B₁ b₁), (B₁ b₂), (B₂ b₁), (B₂ b₂), (B₃ b₁), (B₃ b₂)

• Choices available for black tops = 3
• Choices available for blue lowers = 2

Total number of ways: 3 × 2 = 6